Explanation:
The given chemical reaction is:
The relation between Eo cell and Keq is shown below:
The value of Eo cell is:
Br- undergoes oxidation and I2 undergoes reduction.
Reduction takes place at cathode.
Oxidation takes place at anode.
Hence,
F=96485 C/mol
n=2 mol
R=8.314 J.K-1.mol-1
T=298K
Substitute all these values in the above formula:
Answer:
Keq=6.13x10^33
<span>a.655 k not 100 percent on this but try it. You will use 273.15 and add your Celcius temp to get it in Kelvin
</span>
valence electrons are the number of electrons in the outer shell. there can only be 8 electrons in the outer shell. The number of valence electrons can be used to determine how many bonds are needed.
For example: H2O
O (oxygen) has 6 valence electrons
H (hydrogen) has 1 valence electron
O needs 2 more electrons to be stable
H needs 1 more electron to be stable
O forms one bond with two H atoms to form H2O.
Answer:
310.53 g of Cu.
Explanation:
The balanced equation for the reaction is given below:
CuSO₄ + Zn —> ZnSO₄ + Cu
Next, we shall determine the mass of CuSO₄ that reacted and the mass Cu produced from the balanced equation. This can be obtained as follow:
Molar mass of CuSO₄ = 63.5 + 32 + (16×4)
= 63.5 + 32 + 64
= 159.5 g/mol
Mass of CuSO₄ from the balanced equation = 1 × 159.5 = 159.5 g
Molar mass of Cu = 63.5 g/mol
Mass of Cu from the balanced equation = 1 × 63.5 = 63.5 g
Summary:
From the balanced equation above,
159.5 g of CuSO₄ reacted to produce 63.5 g of Cu.
Finally, we shall determine the mass of Cu produced by the reaction of 780 g of CuSO₄. This can be obtained as follow:
From the balanced equation above,
159.5 g of CuSO₄ reacted to produce 63.5 g of Cu.
Therefore, 780 g of CuSO₄ will react to produce = (780 × 63.5)/159.5 = 310.53 g of Cu.
Thus, 310.53 g of Cu were obtained from the reaction.
Answer:
k= 1.925×10^-4 s^-1
1.2 ×10^20 atoms/s
Explanation:
From the information provided;
t1/2=Half life= 1.00 hour or 3600 seconds
Then;
t1/2= 0.693/k
Where k= rate constant
k= 0.693/t1/2 = 0.693/3600
k= 1.925×10^-4 s^-1
Since 1 mole of the nuclide contains 6.02×10^23 atoms
Rate of decay= rate constant × number of atoms
Rate of decay = 1.925×10^-4 s^-1 ×6.02×10^23 atoms
Rate of decay= 1.2 ×10^20 atoms/s
